What is the area of the region between the graphs of $f(x)=2x^2+5x$ and $g(x)=-x^2-6x+4$ from $x=-4$ to $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{355}{12}$ (Choice B) B $40$ (Choice C) C $\dfrac{128}{3}$ (Choice D) D $8$
Visualizing the area We sketch the graphs of $f$ and $g$ first. ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ ${\llap{-}4}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=-4$ and $x=0$. From this we are looking to evaluate: $ \int_{-4}^{0}\left( g(x)-f(x) \right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{-4}^{0} \left( -x^2-6x+4- (2x^2+5x) \right) \,dx \\\\ &= \int_{-4}^{0} \left( -3x^2-11x+4 \right) \,dx\\\\ &= -x^3-\dfrac{11}{2}x^2+4x~\Bigg|_{-4}^{0} \\\\ &= \left( 0-0+0 \right) -\left( 64-88-16 \right)\\\\ &= 40 \end{aligned}$ Answer The area is $40$ square units.